The mass of sucrose to be added to 300 g of water to lower its vapour pressure by 1.0 mmHg at 25°C is (Given: $\left.p^{\ast }\left(\text{ water }\right)=23.8\mathrm{mmHg}\right)$

Since. $p=x_1{p}_1$ we get

$x_1=\frac{p}{p_1^{\ast }}=\frac{23.8-1.0}{23.8}=0.958$ Now $x_1=\frac{n_1}{n_1+n_2}=\frac{(300/18)\mathrm{mol}}{(300/18)\mathrm{mol}+\left(\mathrm{m}/342\mathrm{g}{\mathrm{mol}}^{-1}\right)}$

Equating these two, we get

$\frac{(300/18)}{(300/18)+(m/342g)}=0.958$ or $m=\frac{\left\{\right(300/18\left)\right(1-0.958\left)\right\}\left(342\mathrm{g}\right)}{0.958}=249.89\mathrm{g}$